Answer:
![62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56) } \, dy = 7875 lb](https://tex.z-dn.net/?f=62.5%5Cint%5Climits%5E6_0%20%7B%28%5Cfrac%7B7%7D%7B6%7Dy%5E%7B2%7D-%5Cfrac%7B49%7D%7B3%7Dy%2B56%29%20%20%7D%20%5C%2C%20dy%20%3D%207875%20lb)
Explanation:
For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).
With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:
γhxdy
we know that γ=62.5 lb/![ft^{3}](https://tex.z-dn.net/?f=ft%5E%7B3%7D)
from the drawing, we can determine the height (or depth under the water) of each differential area is given by:
h=8-y
x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:
![m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7By_%7B2%7D-y_%7B1%7D%7D%7Bx_%7B2%7D-x%7B1%7D%7D)
when substituting the x and y-values given on the graph, we get that the slope is:
![m=\frac{0-6}{7-0}=-\frac{6}{7}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B0-6%7D%7B7-0%7D%3D-%5Cfrac%7B6%7D%7B7%7D)
once we got this slope, we can substitute it in the point-slope form of the equation:
![y_{2}-y_{1}=m(x_{2}-x_{1})](https://tex.z-dn.net/?f=y_%7B2%7D-y_%7B1%7D%3Dm%28x_%7B2%7D-x_%7B1%7D%29)
which yields:
![y-6=-\frac{6}{7}(x-0)](https://tex.z-dn.net/?f=%20y-6%3D-%5Cfrac%7B6%7D%7B7%7D%28x-0%29)
which simplifies to:
![y-6=-\frac{6}{7}x](https://tex.z-dn.net/?f=%20y-6%3D-%5Cfrac%7B6%7D%7B7%7Dx)
we can now solve this equation for x, so we get that:
![x=-\frac{7}{6}y+7](https://tex.z-dn.net/?f=x%3D-%5Cfrac%7B7%7D%7B6%7Dy%2B7)
with this last equation, we can substitute everything into our integral, so it will now look like this:
![\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy](https://tex.z-dn.net/?f=%5Cint%5Climits%5E6_0%7B%2862.5%29%288-y%29%28-%5Cfrac%7B7%7D%7B6%7Dy%2B7%29%7D%5C%2Cdy)
Now that it's all written in terms of y we can now simplify it, so we get:
![62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy](https://tex.z-dn.net/?f=62.5%5Cint%5Climits%5E6_0%20%7B%28%5Cfrac%7B7%7D%7B6%7Dy%5E%7B2%7D-%5Cfrac%7B49%7D%7B3%7Dy%2B56%29%7Ddy%20)
we can now proceed and evaluate it.
When using the power rule on each of the terms, we get the integral to be:
![62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}](https://tex.z-dn.net/?f=62.5%5B%5Cfrac%7B7%7D%7B18%7Dy%5E%7B3%7D-%5Cfrac%7B49%7D%7B6%7Dy%5E%7B2%7D%2B56y%5D%5E%7B6%7D_%7B0%7D)
By using the fundamental theorem of calculus we get:
![62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]](https://tex.z-dn.net/?f=62.5%5B%28%5Cfrac%7B7%7D%7B18%7D%286%29%5E%7B3%7D-%5Cfrac%7B49%7D%7B6%7D%286%29%5E%7B2%7D%2B56%286%29%29-%28%5Cfrac%7B7%7D%7B18%7D%280%29%5E%7B3%7D-%5Cfrac%7B49%7D%7B6%7D%280%29%5E%7B2%7D%2B56%280%29%29%5D)
When solving we get:
![62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56) } \, dy = 7875 lb](https://tex.z-dn.net/?f=62.5%5Cint%5Climits%5E6_0%20%7B%28%5Cfrac%7B7%7D%7B6%7Dy%5E%7B2%7D-%5Cfrac%7B49%7D%7B3%7Dy%2B56%29%20%20%7D%20%5C%2C%20dy%20%3D%207875%20lb)